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MathGroup Archive 1998

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Re: Inverse error function for arguments close to 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg12675] Re: [mg12653] Inverse error function for arguments close to 1
  • From: Wouter Meeussen <eu000949 at pophost.eunet.be>
  • Date: Mon, 1 Jun 1998 23:16:48 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

hi Jacek,

I see only two ways:

1/


Use InverseErfc[arg] for arguments close to 0 :

In[31]:=
Table[N at InverseErf[1-Exp[-x]],{x,0,50}] Out[31]=
{0,0.636716,1.05599,1.38719,1.66819,1.91592,2.13965,2.34504,2.53586,2.71476,
 
2.88367,3.04407,3.19709,3.34365,3.48449,3.62023,3.75136,3.87832,4.00148,
 
4.12116,4.23763,4.35113,4.46188,4.57006,4.67585,4.77939,4.88081,4.98025,
  5.07782,5.17361,5.26769,5.36024,5.45126,5.54067,5.63142,5.711,5.80502,
  5.86358,\[Infinity],\[Infinity],\[Infinity],\[Infinity],\[Infinity],
 
\[Infinity],\[Infinity],\[Infinity],\[Infinity],\[Infinity],\[Infinity],
  \[Infinity],\[Infinity]}
In[32]:=
Table[N at InverseErfc[ Exp[-x]],{x,0,50}] Out[32]=
{0.,0.636716,1.05599,1.38719,1.66819,1.91592,2.13965,2.34504,2.53586,2.71476,
 
2.88367,3.04407,3.19709,3.34365,3.48449,3.62023,3.75136,3.87832,4.00148,
 
4.12116,4.23763,4.35113,4.46188,4.57006,4.67585,4.77938,4.88081,4.98025,
  5.07782,5.1736,5.26771,5.36022,5.4512,5.54074,5.6289,5.71574,5.80131,
  5.88567,5.96888,6.05097,6.13198,6.21197,6.29097,6.369,6.44612,6.52234,
  6.59771,6.67224,6.74596,6.8189,6.89109}

2/

Use "high precision" in stead of machine precision (one of Mathematica
strong points!)

In[26]:=
 Table[{N[InverseErf[1-Exp[-x]],18],
        N[InverseErfc[Exp[-x]],18]},{x,50,60}] 

Out[26]=  
{{6.8910913629004655,6.8910913629004660},
 {6.9625447358489660,6.9625447358489663},
 {7.0332863431220254,7.0332863431220248},
 {7.1033369139523046,7.1033369139523048},
 ...

enjoy,

wouter.


At 17:36 30-05-98 -0400, Jacek Pliszka wrote:
>Hi!
>
>I need to calculate:
>
>InverseErf[1-Exp[-x]] for x \approx 50.
>
>I am trying to use mathematica but it doesn't work for such a large
>values.
>
>Does anybody know how to solve that? Where to look for the proper
>approximation?
>
>I've looked through a few textbooks but I didn't find an answer. The
>only solution I have now is numerical solution of the equation
>Erf[z]+Exp[-x]=1 with high precision calculation for each integral
>needed for iteration step.
>
>Regards,
>
>Jacek Pliszka
>
>Tmp: pliszka at bethe.ucdavis.edu  Perm: pliszka at fuw.edu.pl       
>
>
>
Dr. Wouter L. J. MEEUSSEN
w.meeussen.vdmcc at vandemoortele.be
eu000949 at pophost.eunet.be



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